Question

The temperature at a point (x, y) is T(x, y), measured in degrees Celsius. A bug crawls so that its position after t seconds is given by x = 2 + t , y = 3 + 1 2 t, where x and y are measured in centimeters. The temperature function satisfies Tx(2, 4) = 8 and Ty(2, 4) = 9. How fast is the temperature rising on the bug's path after 2 seconds? (Round your answer to two decimal places.)

Answer 1

Answer:

The temperature is increasing at the rate of 6.5 degree celcius per second.

Step-by-step explanation:

We are given the following information:

Temperature of a point is given by T(x,y)

After t seconds a bug reaches a point $x = \sqrt{2 + t}, y = 3 + \frac{1}{2}t$

At t = 2, x = 2, y = 4

The temperature function satisfies:

$T_x(2,4) = 8\\T_y(2,4) = 9$

We have to find:

$\displaystyle\frac{dT}{dt} = T_t(x(t), y(t))\\\\= T_x(x(t),y(t)).x'(t) + T_y(x(t),y(t)).y'(t) \\\\= T_x(x(t),y(t)).\displaystyle\frac{1}{2\sqrt{(t+2)}} + T_y(x(t),y(t)).\displaystyle\frac{1}{2} \\\\\displaystyle\frac{dT(x,y)}{dt}_{at~t=2} = \displaystyle\frac{dT(2,4)}{dt}_{at~t=2} = 8.\displaystyle\frac{1}{4} + 9.\displaystyle\frac{1}{2} = 2 + 4.5 = 6.5$

Hence, the temperature is increasing at the rate of 6.5 degree celcius per second.